SOLUTION: there are 4 men and 4 women to be seated in a row, and no 2 men or 2 women can sit next to each other. How many combinations of seating are there?

Question 156301: there are 4 men and 4 women to be seated in a row, and no 2 men or 2 women can sit next to each other. How many combinations of seating are there?
Found 2 solutions by gonzo, Ali-aithzaz:
Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
not real sure but this is what i think.
no two men or two women can sit side by side so only can have
mwmwmwmw
or
wmwmwmwm
in the first set (mwmwmwmw) men can arrange themselves in 24 combinations and women can arrange themselves in 24 combinations so total combinations possible are 24 * 24 = 576.
in the second set (wmwmwmwm) same thing i.e. 24 * 24 = 576 possible combinations.
total number of combinations is sum of both sets = 1152.
worked this out with a smaller set (mwmw and mwmwmw) and it looks like it's correct.
combinations of men or women calculated as follows:
in the first woman position any of the 4 women can be in there.
in the second woman position any of the remaining 3 women can be in there.
in the 3d woman position any of the remaining 2 women can be in there.
in the 4th woman position the remaining 1 woman can be in there.
total possible combinations are then 4 * 3 * 2 * 1 = 24.

Answer by Ali-aithzaz(4) (Show Source): You can put this solution on YOUR website!
i think there are two ways they are MWMWMWMW or WMWMWMWM so total ways that no two men an women can't sit together are 2 [ 4! * 4! ] = 1152
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